In fig. (6), the spring has a force constant k The pulley s light and smooth, the spring and string are light. The suspended block has a mass m. If the block is slightly displaced from its equilibrium position and then released, the period of its vertical oscillation is

see fig (12)Let the tension in spring at equilibrium position for block beT0. Then mg=T0 From force diagram of pulley2T0=ky0  or  2mg=ky0 When the block is displaced through a small distanceyin downward direction from equilibrium position, let the tension in string be T. The stretching of string will be (y/2). Therefore  and 2T=ky0+y02=ky0+ky02 From eqs. (1) and 2), we get 2T=2mg+ky02  or  T=mg+ky04 Now mg−mg+ky04=ma or ma=−ky04  or  a=−k4my0 therofore d2y0dt2=−k4my0  or  d2y0dt2+k4my0=0 T=2πω=2π4mR